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A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at $14 \mathrm{~cm}$ from the centre of the magnet. The earth's magnetic field at the place is $0.36 \mathrm{G}$ and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null-point (i.e., $14 \mathrm{~cm}$ ) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field).
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Verified Answer
Given: Angle of dip $=0^{\circ}$; Distance of null point $=14 \mathrm{~cm}$ Earth's magnetic field $\mathrm{B}=0.36 \mathrm{G}$
To find: Total magnetic field on the normal bisector of the magnet $\mathrm{B}^{\prime}$
Formula used: $\mathrm{B}=\mathrm{B}_{\text {axial }}+\mathrm{B}_{\text {equatorial }}$
In case of bar magnet placed in horizontal plane in N-S direction, null points wil lie along the axial line.
At the null point. H (Horizontal component of earth's field) is given by
$$
\mathrm{H}=\mathrm{B}_{\text {axial }} \text { and } \mathrm{B}_{\text {axial }}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{r}^3}=\mathrm{H}
$$
Also $\mathrm{B}_{\mathrm{eq}}=\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}^3} \Rightarrow \mathrm{B}_{\mathrm{eq}}=\frac{\mathrm{H}}{2}$
Hence, total magnetic field of magnet
$$
\begin{aligned}
&\mathrm{B}=\mathrm{B}_{\mathrm{axial}+} \mathrm{B}_{\mathrm{eq}}=\mathrm{H}+\frac{\mathrm{H}}{2}=\frac{3}{2} \mathrm{H}=\frac{3}{2} \times 0.36 \\
&\Rightarrow \mathrm{B}=0.54 \mathrm{G}
\end{aligned}
$$
To find: Total magnetic field on the normal bisector of the magnet $\mathrm{B}^{\prime}$
Formula used: $\mathrm{B}=\mathrm{B}_{\text {axial }}+\mathrm{B}_{\text {equatorial }}$
In case of bar magnet placed in horizontal plane in N-S direction, null points wil lie along the axial line.
At the null point. H (Horizontal component of earth's field) is given by
$$
\mathrm{H}=\mathrm{B}_{\text {axial }} \text { and } \mathrm{B}_{\text {axial }}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{r}^3}=\mathrm{H}
$$
Also $\mathrm{B}_{\mathrm{eq}}=\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}^3} \Rightarrow \mathrm{B}_{\mathrm{eq}}=\frac{\mathrm{H}}{2}$
Hence, total magnetic field of magnet
$$
\begin{aligned}
&\mathrm{B}=\mathrm{B}_{\mathrm{axial}+} \mathrm{B}_{\mathrm{eq}}=\mathrm{H}+\frac{\mathrm{H}}{2}=\frac{3}{2} \mathrm{H}=\frac{3}{2} \times 0.36 \\
&\Rightarrow \mathrm{B}=0.54 \mathrm{G}
\end{aligned}
$$
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